Answer to Question #165477 in Mechanical Engineering for Rodel

Question

Ammola vapor is compressed in a closed system from 15 psia, 3 cu ft, to 150 psia, 0.4 cu ft. according to the relation pVⁿ = constant. Assumin the vapor to be an ideal gas with the gas constant as 90.72 lb_m-ft/lb_f -°R and c_v = 0.384 BTU/lb_m - °R compute the heat transferred and the work.

Answer 1

As per the relation

$PV^n= C, P_1V_1^n=P_2V_2^n$

$15\times 3^n= 150\times 0.4^n$

$(\frac{3}{0.4})^n=10$

take log both sides

$nlog(7.5)= log(10)$

0.875n=1

n=1.14

so here value of n which follows above equation is 1.14

Now, we know that work done for polytropic process is

$W=\frac{P_1V_1-P_2V_2}{n-1}=-107.14$

from ideal gas equation we can find value of temperature as

$PV=nRT, T_1=\frac{P_1V_1}{nR}=\frac{15 \times 3}{1\times 90.72}$ =496 K,$T_2= \frac{P_2V_2}{nR}=661.37 K$

$\Delta U= n C_vdT=1\times 0.384\times 165.37=62.508 ft-lb_f$

Heat transfer as per first law of thermodynamics

$\Delta Q= \Delta U + \Delta W$

$\Delta Q=-43.63 ft lb_f$